Question: Select the unit vector in the direction of $\vec{v}=\left( -8, -6 \right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left( {-\dfrac{4}{5}},{-\dfrac{3}{5}}\right) $ (Choice B) B $\left( {-\dfrac{8}{5}},{-\dfrac{6}{5}}\right) $ (Choice C) C $\left( {-\dfrac{4}{\sqrt{5}}},{-\dfrac{3}{\sqrt{5}}}\right) $ (Choice D) D $\left( {-\dfrac{8}{\sqrt{5}}},{-\dfrac{6}{\sqrt{5}}}\right) $
Solution: Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( -8, -6 \right) }{\sqrt{(-8)^2+(-6)^2}} \\\\\\ &= \dfrac{1}{\sqrt{100}} \cdot \left( -8, -6\right) \\\\\\ &= \left( -\dfrac{8}{10}, -\dfrac{6}{10}\right)\\\\\\ &= \left( {-\dfrac{4}{5}}, {-\dfrac{3}{5}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {-\dfrac{4}{5}} \right)^2 + \left( {-\dfrac{3}{5}} \right)^2} \\\\\\ &= \sqrt{\dfrac{16}{25} + \dfrac{9}{25}} \\\\\\ &= \sqrt{\dfrac{25}{25}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {-\dfrac{4}{5}}, {-\dfrac{3}{5}}\right) $ Visualizing the answer: $-8$ $-6$ $\dfrac{-8}{10}=-\dfrac{4}{5}$ $\dfrac{-6}{10}=-\dfrac{3}{5}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$.